You follow a systematic procedure to balance the equation.
Start with the unbalanced equation:
#"SO"_2 + "O"_2 → "SO"_3#
A method that often works is to balance everything other than #"O"# and #"H"# first, then balance #"O"#, and finally balance #"H"#.
Another useful procedure is to start with what looks like the most complicated formula.
The most complicated formula looks like #"SO"_3#. We put a 1 in front of it to remind ourselves that the number is now fixed.
#"SO"_2 + "O"_2 → color(red)(1) "SO"_3#
Balance #"S"#:
We have #"1 S"# on the right, so we need #"1 S"# on the left. We put a 1 in front of the #"SO"_2#.
#color(blue)(1) "SO"_2 + "O"_2 → color(red)(1) "SO"_3#
Balance #"O"#:
We have #"3 O"# on the right, so we need #"3 O"# on the left. There are already #"4 O"# atoms on the left. We must put a ½ in front of the #"O"_2#.
Most instructors don't allow fractions in balanced chemical equations, because you can't have a fraction of an atom or a molecule.
To get rid of the fractions, you multiply all the coefficients by 2 and get
#color(blue)(2) "SO"_2 + "O"_2 → color(red)(2) "SO"_3#
Now you can balance the equation by putting a 1 in front of the #"O"_2#.
#color(blue)(2) "SO"_2 + color(orange)(1) "O"_2 → color(red)(1) "SO"_3#
We should now have a balanced equation.
Check to make sure:
#bb"Atom"color(white)(m)bb "Left hand side"color(white)(m) bb"Right hand side"#
#stackrel(————————————————)(color(white)(ml)"S"color(white)(mmmmmm) 2color(white)(mmmmmmmll) 2)#
#color(white)(ml)"O"color(white)(mmmmmll) 6color(white)(mmmmmmmll) 6#
The balanced equation is
#"2SO"_2 + "O"_2 → "2SO"_3#
